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2

The "Maximum Principle" is a very useful tool to answer to such a question (as John Barber rightly did). I will not come back with this principle to repeat what was already said and which is more general for all times from $t=0$ to $t\to\infty$. In the present case, since the wording of the question concerns only $t\to\infty$ and where it is not question of ...


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Since $\|f - f_n \|_p \to 0$, we can extract a subsequence $f_{n_j}$ so that $\| f - f_{n_j} \|_p \le \frac{1}{2^j}$. Put $$g = \lvert f \rvert + \sum^\infty_{j=1} \lvert f - f_{n_j} \rvert.$$ Then $g \in L^p$ and by the triangle inequality we have $$\lvert f_{n_j} \rvert \le g \,\,\, (\text{almost everywhere}).$$ From this subsequence, you can extract a ...


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Assume that $v_1^2+v_2^2\neq 0$. Write $u(x,y)=U(z,w)$ with $$z:=v_1x+v_2y\text{ and }w:=v_2x-v_1y\,.$$ Hence, $$x=\frac{v_1z+v_2w}{v_1^2+v_2^2}\text{ and }y=\frac{v_2z-v_1w}{v_1^2+v_2^2}\,.$$ Observe that $z$ and $w$ are independent (by checking that $\frac{\partial z}{\partial w}=0$ and $\frac{\partial w}{\partial z}=0$). Then, you can see that $$\frac{\...


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To be clear, you want to find $P_n(t)$ for a birth-death process with constant birth rate $q(i,i+1)=\lambda$ and constant death rate $q(i,i-1)=\mu$? This particular birth and death process is exactly the $M/M/1$ queue. As you can see, under "Transient solution", there is a solution for the probability mass function dependent on time for a particular state.


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Differentiate both sides of the equation $f(\lambda x) = \lambda^{a}f(x)$ in $\lambda$ to obtain: \begin{equation} \sum_{i = 1}^{n}x_{i}(\partial_{i}f)(\lambda x) = a\lambda^{a-1}f(x) \end{equation} Setting $\lambda = 1$ yields $\sum_{i = 1}^{n}x_{i}(\partial_{i}f)(x) = af(x)$, known as Euler's identity. Differentiating the equation displayed above in $\...


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You can find the general statement and proof in Chapter 6 of the book "Sobolev Spaces" by Robert A. Adams and John. J. F. Fournier.


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Use the Co-area formula with $g=|\nabla u|^{-1} \chi_{\Omega_t}$ to get $$ \int_{\Omega_t} dA=\int_t^\infty \left(\int_{\partial \Omega_x} \frac{ds}{|\nabla u|}\right)dx.$$ Differentiating this with respect to $t$ gives the formula you're looking for.


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Let $L$ be a $k^{th}$-order linear differential operator, i.e. one which satisfies $L(\alpha u + \beta v) = \alpha L u + \beta Lv$ for all $u,v\in C^k$ and constants $\alpha,\beta$ (of course this notion can be weakened past $C^k$, but this will do for here). We say that the equation $L u =f$, for some given function $f$, is linear. A semilinear equation ...


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If we set $$x(\xi, t) = f(\xi + \beta t, t) = f(z, t)$$ we find (by the chain rule) \begin{align} x_{t} &= f_{z} \cdot z_{t} + f_{t} \\ &= \beta f_{z} + f_{t} \\ x_{\xi} &= f_{z} \cdot z_{\xi} \\ &= f_{z} \\ x_{\xi \xi} &= f_{zz} \cdot z_{\xi} \\ &= f_{zz} \end{align} Substituting into our original PDE, we get \begin{align} x_{t}...


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The only way to have a spatially smooth solution is for $\partial u/\partial r$ go to zero as $r$ goes to zero. This means you can use L'Hopitals's rule in evaluating that second term: $\displaystyle \frac{1}{r}\frac{\partial u}{\partial r} \rightarrow \frac{\partial_r(\partial u/\partial r)}{\partial_r(r)} = \frac{\partial^2 u}{\partial r^2}$


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You can transform to Sturm-Liouville form by dividing by $x$: $$ (xM')'+\frac{\lambda}{x}M = 0. $$ Whenever you have such a form, you can get rid of the multiplier in the derivative. There is a standard trick when you change the independent variable. In this case, let $M(x) = P(\int\frac{1}{x}dx)$. Then $$ xM'(x) = ...


1

$$x^2\frac{d^2M}{dx^2}+x\frac{dM}{dx}+\lambda M(x)=0$$ The change of variable $\quad z=e^x\quad\to\quad x=\ln|z[\quad\to\quad \frac{dz}{dx}=z \quad$ isn't a good idea because it leads to a more complicated form of ODE : $\frac{dM}{dx}=\frac{dM}{dz}\frac{dz}{dx}=z\frac{dM}{dz}$ $\frac{d^2M}{dx^2}=\left(\frac{d}{dz}\frac{dM}{dx}\right)\frac{dz}{dx}=\frac{d}{...


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Another particular solution you missed can be found as: $\frac{1}{(D+D')^2}x=\frac{1}{D'^2}$$[1+\frac{D}{D'}]^{-2}x=\frac{1}{D'^2}(1-2D/D'+....]x=\frac{1}{D'^2}(x-2y)=\frac{xy^2}{2}-\frac{y^3}{3}$



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